The mole and stoichiometry

Atoms and molecules are far too small and numerous to count individually, so chemists count them in a fixed package called the mole. One mole contains the Avogadro constant, $N_A = 6.02 \times 10^{23}\ \text{mol}^{-1}$, of particles - a number given in the data booklet.

• A mole is the amount of substance that holds exactly $6.02214076 \times 10^{23}$ specified particles. Since 2019 this count has been fixed by definition as the Avogadro constant, so the mole is now pinned to a counted number rather than to a mass of any element.
• (Before 2019 the mole was instead tied to a sample of carbon-12: it was defined as the number of atoms found in exactly 12 g of that isotope. You may still meet this older wording in textbooks.)
• The mole works for any particle - you must always say which: atoms, molecules, ions, formula units or electrons.

To go between amount (mol) and the actual number of particles:

$$N = n \times N_A$$

For example, $0.50\ \text{mol}$ of helium contains $0.50 \times 6.02 \times 10^{23} = 3.0 \times 10^{23}$ atoms.

Be careful with the particles inside a formula. In $1\ \text{mol}$ of $\text{CO}_2$ there is:

• $1\ \text{mol}$ of $\text{CO}_2$ molecules,
• $1\ \text{mol}$ of C atoms and $2\ \text{mol}$ of O atoms,
• so $3 \times 6.02 \times 10^{23} = 1.8 \times 10^{24}$ atoms in total.

The mole is the bridge between the microscopic world of particles and macroscopic masses you can weigh in the lab.

Masses of single atoms are tiny, so we compare them on a relative scale based on carbon-12.

• Relative atomic mass ($A_r$) is the weighted average mass of an element's atoms relative to $\tfrac{1}{12}$ of a carbon-12 atom. It is a weighted average because most elements are mixtures of isotopes - this is why $A_r(\text{Cl}) = 35.45$, not a whole number.
• Relative molecular mass ($M_r$) applies to molecules and is the sum of the $A_r$ values of all the atoms in the formula.
• Relative formula mass is the same idea for ionic or giant structures (which have no molecules), e.g. for $\text{NaCl}$.

All of these are ratios, so they have no units.

Worked example - $M_r$ of calcium hydroxide, $\text{Ca(OH)}_2$:

$$M_r = 40.08 + 2(16.00 + 1.01) = 74.10$$

Remember the subscript multiplies everything inside the brackets. The $A_r$ values you need are taken from the periodic table in the data booklet - use the values given there, not rounded whole numbers, unless the question allows it.

The molar mass $M$ is the mass of one mole of a substance, in $\text{g mol}^{-1}$. Conveniently, it is numerically equal to the relative atomic or formula mass:

• $A_r(\text{Na}) = 22.99 \Rightarrow M = 22.99\ \text{g mol}^{-1}$
• $M_r(\text{H}_2\text{O}) = 18.02 \Rightarrow M = 18.02\ \text{g mol}^{-1}$

The key conversion, given in the data booklet, is:

$$n = \frac{m}{M}$$

where $n$ is amount (mol), $m$ is mass (g) and $M$ is molar mass ($\text{g mol}^{-1}$). Rearrange as needed: $m = nM$ or $M = m/n$.

This equation is the heart of nearly every stoichiometry calculation. A reliable routine is:

• convert each known mass to moles with $n = m/M$;
• use the mole ratio from the balanced equation to find moles of the species you want;
• convert back to mass, volume or concentration as required.

For example, $9.0\ \text{g}$ of water is

$$n = \frac{9.0}{18.02} = 0.50\ \text{mol}.$$

• The empirical formula is the simplest whole-number ratio of atoms of each element.
• The molecular formula is the actual number of each atom in a molecule - always a whole-number multiple of the empirical formula.

Finding the empirical formula from masses or percentages:

1. Write the mass (or % by mass) of each element.

2. Divide each by its $A_r$ to get moles of atoms.

3. Divide all by the smallest of these mole values.

4. Multiply up to the nearest whole numbers if needed (e.g. $1.5 \to 3$).

Finding the molecular formula: divide the (relative) molar mass by the empirical formula mass to get the multiplier $x$, then multiply every subscript by $x$.

$$x = \frac{M_r(\text{molecule})}{M_r(\text{empirical unit})}$$

Example: a compound is $\text{CH}_2$ empirically with empirical mass $14.03$. If $M_r = 56.0$, then $x = 56.0/14.03 \approx 4$, so the molecular formula is $\text{C}_4\text{H}_8$.

With combustion or hydrate data, work out the mass of each element (or water) first, then follow the same mole-ratio steps.

Molar concentration is the amount of solute dissolved per cubic decimetre of solution, in $\text{mol dm}^{-3}$. Square brackets denote concentration, e.g. $[\text{NaOH}] = 0.10\ \text{mol dm}^{-3}$.

The data booklet gives:

$$n = CV$$

with $C$ in $\text{mol dm}^{-3}$ and $V$ in $\text{dm}^3$. Rearrange to find concentration:

$$C = \frac{n}{V}$$

Unit care is essential:

• $1\ \text{dm}^3 = 1000\ \text{cm}^3$, so divide a $\text{cm}^3$ volume by 1000 before using it.
• $1\ \text{dm}^3 = 1\ \text{L}$.

Example: dissolving $0.20\ \text{mol}$ of solute in $500\ \text{cm}^3$ ($= 0.500\ \text{dm}^3$) gives

$$C = \frac{0.20}{0.500} = 0.40\ \text{mol dm}^{-3}.$$

For dilution, the moles of solute are unchanged, so $C_1V_1 = C_2V_2$ (a handy consequence of $n = CV$). Concentration can also be expressed in $\text{g dm}^{-3}$; convert to $\text{mol dm}^{-3}$ by dividing by the molar mass.

Avogadro's law tells us that, at a fixed temperature and pressure, any two gas samples of the same volume hold the same number of molecules - and therefore the same number of moles. The practical consequence is powerful:

• For reacting gases, the volume ratio equals the mole ratio from the balanced equation. No molar masses are needed.

Example: $\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$. At the same T and P, $10\ \text{cm}^3$ of $\text{N}_2$ reacts with $30\ \text{cm}^3$ of $\text{H}_2$ to give $20\ \text{cm}^3$ of $\text{NH}_3$.

To connect a gas volume to moles, use the molar volume at STP (273.15 K, 100 kPa), given in the data booklet as $V_m = 22.7\ \text{dm}^3\,\text{mol}^{-1}$:

$$n = \frac{V}{V_m}$$

• Use $22.7$, not $24.0$ - the latter is the older RTP value.
• $V$ must be in $\text{dm}^3$.

Example: $4.54\ \text{dm}^3$ of any gas at STP is

$$n = \frac{4.54}{22.7} = 0.200\ \text{mol}.$$

For gases not at STP, the ideal gas equation $PV = nRT$ (topic S1.5) is used instead.